Thevinin's theorem relies on simplifying a circuit such that you remove the element for which you are trying to analyze its current/voltage drop and determine the equivalent voltage and in-line resistance of the remaining circuit. This gives you a series resistor and voltage source that are equivalent in behavior to the complex circuit pictured.

To calculate the voltage, you treat the 10 ohm resistor like an open circuit and determine the voltage between those points. Tahis requires us to use some equations based on Kirchoff's voltage law (KVL) and current law (KCL). I'll base it on the current flowing through the one-ohm resistor in a counter-clockwise direction (`i_1`), the current flowing downward through the 5-ohm resistor (`i_2`), and the current flowing in a clockwise direction through the 2-ohm resistor (`i_3`). From the Kirchoff laws, we get 3 equations:

KVL:

`20-i_1-5i_2+12=0`

`10-2i_3-5i_2+12=0`

KCL:

`i_1-i_2+i_3=0`

Solving these for `i_2` obtains 86/17 or 5.06 A. This means that the voltage drop across the 5 ohm resistor is 5.06*5 = 25.3 V. To get the final Thevinin voltage, we sum this 25.3 V with the -12 V from the source in series with the resistor to obtain a **final Thevinin voltage of 13.3 V.**

To obtain the Thevinin resistance, you use the same two terminals that connect to the 10 ohm resistor, but then you short all of the voltage sources and open each current source. By doing that in this case, you end up with 1, 2, and 5 ohm resistors in parallel. Calculaing the overall resistance:

`1/(1/1+1/2+1/5)=1/1.7=0.59 Omega`

Thus, we have our series resistance: 0.59 ohms.

We can now make our Thevinin circuit with the voltage source of 13.3 V in series with the source resistance of 0.59 ohms, now in series with the 10 ohm resistor. Simplifying the resistance in this case is easy because they are in series: we add them such that the overall resistance is 10.59 ohms. Calculating the current using Ohm's law gives us the final answer:

`(13.3V)/(10.59 Omega) = 1.26 A`

**Further Reading**

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